lambda converted to bool instead of deducing function-pointer-type - c++

Question

I wanted to implement a overload for operator<< that allowed me to call a given function and output the result.
I therefore wrote an overload, but the conversion to bool is selected and when writing a function myself, it would not compile.
EDIT: Know that I do not want to call the lambda,
but instead pass it to the function where it should be called with a default constructed parameter list.
I have appended my code:
#include <iostream>
template<typename T>
void test(T *) {
std::cout << "ptr" << std::endl;
}
template<typename T>
void test(bool) {
std::cout << "bool" << std::endl;
}
template<typename Ret, typename ...Args>
void test(Ret(*el)(Args...)) {
std::cout << "function ptr\n" << el(Args()...) << std::endl;
}
template<typename Char_T, typename Char_Traits, typename Ret, typename ...Args>
std::basic_ostream<Char_T, Char_Traits>& operator<<(
std::basic_ostream<Char_T, Char_Traits> &str, Ret(*el)(Args...)) {
return str << el(Args()...);
}
int main() {
std::boolalpha(std::cout);
std::cout << []{return 5;} << std::endl; // true is outputted
test([]{return 5;}); // will not compile
}
I use gcc 7.3.1 with the version flag -std=c++14.
EDIT: Error message:
main.cc: In function ‘int main()’:
main.cc:25:23: error: no matching function for call to ‘test(main()::<lambda()>)’
test([]{return 5;});
^
main.cc:5:6: note: candidate: template<class T> void test(T*)
void test(T *) {
^~~~
main.cc:5:6: note: template argument deduction/substitution failed:
main.cc:25:23: note: mismatched types ‘T*’ and ‘main()::<lambda()>’
test([]{return 5;});
^
main.cc:9:6: note: candidate: template<class T> void test(bool)
void test(bool) {
^~~~
main.cc:9:6: note: template argument deduction/substitution failed:
main.cc:25:23: note: couldn't deduce template parameter ‘T’
test([]{return 5;});
^
main.cc:13:6: note: candidate: template<class Ret, class ... Args> void test(Ret (*)(Args ...))
void test(Ret(*el)(Args...)) {
^~~~
main.cc:13:6: note: template argument deduction/substitution failed:
main.cc:25:23: note: mismatched types ‘Ret (*)(Args ...)’ and ‘main()::<lambda()>’
test([]{return 5;});

Answer 1

Your problem here is that Template Argument Deduction is only done on the actual argument passed to test. It's not done on all possible types that the argument could possibly converted to. That might be an infinite set, so that's clearly a no-go.
So, Template Argument Deduction is done on the actual lambda object, which has an unspeakable class type. So the deduction for test(T*) fails as the lambda object is not a pointer. T can't be deduced from test(bool), obviously. Finally, the deduction fails for test(Ret(*el)(Args...)) as the lambda object is not a pointer-to-function either.
There are a few options. You might not even need a template, you could accept a std::function<void(void)> and rely on the fact that it has a templated constructor. Or you could just take a test(T t) argument and call it as t(). T will now deduce to the actual lambda type. The most fancy solution is probably using std::invoke, and accepting a template vararg list.

Answer 2

Even though non-capturing lambdas have an implicit conversion to function pointers, function templates must match exactly for deduction to succeed, no conversions will be performed.
Therefore the easiest fix is to force the conversion with a +
int main() {
std::boolalpha(std::cout);
std::cout << []{return 5;} << std::endl; // true is outputted
test(+[]{return 5;});
// ^
}

Answer 3

template<typename T>
void test(bool) {
std::cout << "bool" << std::endl;
}
Template is not needed. In fact you overload functions, not templates. Replace it with
void test(bool) {
std::cout << "bool" << std::endl;
}
Now your sample will compile.